# Deriving the Formula of the Vertex of Quadratic Functions

In getting the vertex of the quadratic function in general form $f(x) = ax^2 + bx + c$, we usually need to convert it to the vertex form $f(x) = a(x - h)^2 + k$. In the latter form, the vertex of the parabola is at $(h,k)$. For example, the function in the general form

$f(x) = 2x^2 - 12x + 22$

can be rewritten in the vertex form as

$f(x) = 2(x - 3)^2 + 4$.

In the vertex form, it is easy to see that the vertex is at $(h,k) = (3,4)$.

Aside from this method, we can also use the ordered pair

$\left ( \dfrac{-b}{2a}, \dfrac{4ac - b^2}{4a} \right )$

in place of $(h,k)$

This means that we can get the vertex of $f$ without changing the general form to vertex from. In this post, we are going to derive the ordered pair above. That is, we are going to show that

$h = \dfrac{-b}{2a}$

and

$k = \dfrac{4ac - b^2}{4a}$.

Expanding $f(x) = a(x - h)^2 + k$ gives us

$f(x) = a(x^2 - 2hx + h^2) + k$
$f(x) = ax^2 - 2ahx + ah^2 + k$.

Now, the terms with both x’s in the general and vertex forms are $bx$ and $-2ahx$. So, we can equate the two expressions and solve for $h$.

$-2ahx = bx$
$-2ah = b$
$h = \frac{- b}{2a}$.

Now, we can use $h$ as the value of $x$ in the general from. That is

$f(h) = ah^2 + bh + c$
$f\left ( \dfrac{-b}{2a} \right ) = a \left ( \dfrac{-b}{2a} \right )^2 + b \left (\dfrac{-b}{2a} \right ) + c$
$= a \left (\dfrac{b^2}{4a^2} \right ) - \dfrac{-b^2}{2a} + c$
$= \dfrac{b^2}{4a} - \dfrac{b^2}{2a} + c$
$= \dfrac{b^2 - 2b^2 + 4ac}{4a}$
$= \dfrac{4ac - b^2}{4a}$
$k = \dfrac{4ac - b^2}{4a}$ .

This means that we can get the vertex of the function

$f(x) = 2x^2 - 12x + 22$

without converting it to the vertex form as follows.

$h = \dfrac{-b}{2a} = \dfrac{-(-12)}{2(2)} = 3$
$k = \dfrac{4(2)(22) - (-12)^2}{4(2)} = \dfrac{32}{8} = 4$

So, the vertex is at $(h,k) = (3,4)$ which is consistent with our calculation above.